If $$f(x)=\int_{}^{}\frac{1}{x^{1/4}(1+x^{1/4})}dx,f(0)=-6$$, then $$f(1)$$ is equal to :

To solve the problem, we start by evaluating the indefinite integral:

Given: $$f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} dx$$

We use the substitution $$t = x^{1/4}$$. Then, $$dt = \frac{1}{4} x^{-3/4} dx$$, so $$dx = 4 x^{3/4} dt$$.

Since $$t = x^{1/4}$$, we have $$x^{3/4} = (x^{1/4})^3 = t^3$$. Thus, $$dx = 4 t^3 dt$$.

Substitute into the integral:

$$\int \frac{1}{x^{1/4} (1 + x^{1/4})} dx = \int \frac{1}{t (1 + t)} \cdot 4 t^3 dt = \int \frac{4 t^3}{t (1 + t)} dt = \int \frac{4 t^2}{1 + t} dt$$

Simplify the integrand by polynomial division:

$$\frac{t^2}{t + 1} = t - 1 + \frac{1}{t + 1}$$

So, $$\frac{4 t^2}{1 + t} = 4 \left( t - 1 + \frac{1}{t + 1} \right) = 4t - 4 + \frac{4}{t + 1}$$

Integrate:

$$\int \left( 4t - 4 + \frac{4}{t + 1} \right) dt = 4 \cdot \frac{t^2}{2} - 4t + 4 \ln |t + 1| + C = 2 t^2 - 4t + 4 \ln |t + 1| + C$$

Substitute back $$t = x^{1/4}$$:

$$f(x) = 2 (x^{1/4})^2 - 4 (x^{1/4}) + 4 \ln |x^{1/4} + 1| + C = 2 \sqrt{x} - 4 x^{1/4} + 4 \ln (x^{1/4} + 1) + C$$

Since $$x^{1/4} \geq 0$$ for $$x \geq 0$$, the absolute value can be removed:

$$f(x) = 2 \sqrt{x} - 4 x^{1/4} + 4 \ln (x^{1/4} + 1) + C$$

Given $$f(0) = -6$$, evaluate the limit as $$x \to 0^+$$:

$$\lim_{x \to 0^+} \left[ 2 \sqrt{x} - 4 x^{1/4} + 4 \ln (x^{1/4} + 1) + C \right] = 2(0) - 4(0) + 4 \ln (0 + 1) + C = 4 \ln 1 + C = 4 \cdot 0 + C = C$$

Set equal to the given condition: $$C = -6$$

Thus, $$f(x) = 2 \sqrt{x} - 4 x^{1/4} + 4 \ln (x^{1/4} + 1) - 6$$

Now find $$f(1)$$:

$$f(1) = 2 \sqrt{1} - 4 (1)^{1/4} + 4 \ln (1^{1/4} + 1) - 6 = 2(1) - 4(1) + 4 \ln (1 + 1) - 6 = 2 - 4 + 4 \ln 2 - 6 = -2 + 4 \ln 2 - 6 = 4 \ln 2 - 8$$

Factor: $$4 \ln 2 - 8 = 4 (\ln 2 - 2)$$

Comparing with the options:

A. $$4(\log_{e}2-2) = 4(\ln 2 - 2)$$
B. $$2 - \log_{e^{2}}2$$ (which simplifies to $$2 - \frac{\ln 2}{2}$$, not matching)
C. $$\log_{e}2 + 2 = \ln 2 + 2$$, not matching
D. $$4(\log_{e}2+2) = 4(\ln 2 + 2)$$, not matching

Thus, the correct option is A.